JEE Binomial Theorem Previous Year Questions With Solutions

The binomial theorem describes the coefficients of each term when we expand (a + b)ⁿ. When n is equal to a prime number p, p will divide all those coefficients, except the end one’s. Binomial theorem is one of the important concepts in JEE Mathematics. This theorem also plays a prominent role to proof other results or theorems. About 1-2 question/s asked from this topic in JEE Examination. The subject experts at BYJU’S bring chapter-wise previous year solved questions of Binomial Theorem including important concepts and formulae to help JEE aspirants. The study material given in this article can be used for a quick revision before the examination. Students can freely download the solutions and start practising offline.

Download Binomial Theorem Previous Year Solved Questions PDF

 Important concepts to Remember:

Results for Specific Types of Binomial Expansions

JEE Main Past Year Questions With Solutions on Binomial Theorem

Question 1: If the sum of the coefficients of all even powers of x in the product (1 + x + x² + … + x²ⁿ) (1 – x + x² – x³ + … + x²ⁿ) is 61, then find the value of n.

Solution:

Let (1 + x + x² + … + x²ⁿ) (1 – x + x² – x³ + … + x²ⁿ) = a₀ + a₁x + a₂x² + a₃x³+ ……

Put x = 1 => 2n + 1 = a₀ + a₁ + a₂ + a₃ + …… …(i)

Put x = -1 => 2n + 1 = a₀ – a₁ + a₂ – a₃ + …… …..(ii)

Add (i) and (ii), we get

2n + 1 = 61

Or n = 30

Question 2: If α and β be the coefficients of x⁴ and x² respectively in the expansion of (x + √(x² − 1))⁶ + (x − √(x² − 1)⁶, then:

(a) α + β = −30

(b) α− β = −132

(c) α− β = 60

(d) α+ β = 60

Answer: (b)

Solution: (x + √(x² − 1))⁶ + (x − √(x² − 1)⁶

= 2[⁶C₀ x⁶ + ⁶C₂x⁴(x² – 1) + ⁶C₄x²(x² – 1)² + ⁶C₆(x² – 1)³]

= 2[32x⁶ – 48x⁴ + 18x² – 1]

=> α = -96 and β = 36

=> α – β = -132

Question 3: Find the coefficient of x⁴ in the expansion of (1 + x + x²)¹⁰ .

Solution:

General term of the given expression is,

Here, q + 2r = 4

For p = 6, q = 4, r = 0, coefficient = 10!/[6! x 4!] = 210

For p = 7, q = 2, r = 1, coefficient = 10! /[7! x 2! x 1!] = 360

For p = 8, q = 0, r = 2, coefficient = 10!/[ 8! x 2!] = 45

Therefore, sum = 615

Question 4: If x is positive, the first negative term in the expansion of

(1 + x)^27/5 is

(a) 5th term (b) 6th term (c) 7th term (d) 8th term

Answer: (d)

Solution:

We know,

Therefore, n – r + 1 < 0

=> 27/5 + 1 < r

or r > 6

Here r = 7

So, T_r+1 < 0

Hence T_r+1 = T₇₊₁ = T₈ = 8th term is negative.

Question 5: The positive integer just greater than (1 + 0.0001)¹⁰⁰⁰⁰ is

(a) 4 (b) 5 (c) 2 (d) 3

Answer: (d)

Solution:

(1 + 0.0001)¹⁰⁰⁰⁰ which is similar to the form (1 + 1/n)ⁿ where n = 10000.

Using binomial expansion, we have

(1 + 0.0001)¹⁰⁰⁰⁰ = (1 + 1/n)ⁿ

= 1 + n × (1/n) + n(n-1)/2! × (1/n² ) + ([n(n-1)(n-2)]/3! )× (1/n³)+ ……..

= 2 + n(n – 1)/2n² + ….

= 2 + a positive quantity

> 2

So just greater positive integer is 3.

Question 6: The coefficients of x^p and x^q in the expression of (1 + x)^p+q are

(a) Equal

(b) Equal with opposite signs

(c) Reciprocals of each other

(d) None of these

Answer: (a)

Solution:

The coefficients of x^p and x^q in the expression of (1 + x)^p+q

We know, y_p+1 = ^p+qC_p x^p and y_q+1 = ^p+qC_q x^q

Coefficient of x^p = ^p+qC_p = [p+q]!/p!q!

and

Coefficient of x^q = ^p+qC_q = [p+q]!/p!q!

Therefore, coefficients of x^p and x^q in the expression of (1 + x)^p+q are equal.

Question 7: If

(a) a_n < 7 for all n ≥ 1

(b) a_n > 7 for all n ≥ 1

(c) a_n > 3 for all n ≥ 1

(d) a_n < 4 for all n ≥ 1

Answer: (c)

Solution:

=>a_n = √(7+a_n)

=> a_n² – a_n – 7 = 0

Solving above quadratic equation, we get

a_n = [1±√29]/2

But a_n > 0, therefore, a_n = [1+√29]/2 > 3

=> a_n > 3

Question 8: If the sum of the coefficients in the expansion of (a + b)ⁿ is 4096, then the greatest coefficient in the expansion is

(a) 1594 (b) 924 (c) 792 (d) 2924

Answer: (b)

Solution:

Binomial expansion: (x + y)ⁿ = C₀ xⁿ + C₁ x^n-1 y + C₂ x^n-2y² + ….+ C_nyⁿ

Putting x = y = 1, we get

2ⁿ = C₀ + C₁ + C₂ + ….+ C_n = 4096 = 2¹²

=> n = 12

Since n is even here, so the coefficient of greatest term is

ⁿC_n/2 = ¹²C_12/2 = ¹²C₆

= (12/6) × (11/5) × (10/4) × (9/3) × (8/2) × (7/1)

= 924

Question 9: The sum of the co-efficient of all odd degree terms in the expansion of

(x + √(x³-1)⁵ + (x – √(x³-1)⁵ , (x > 1) is

(a) 0 (b) 2 (c) 1 (d) -1

Answer: (b)

Solution:

Let y = √(x³-1)

So, given expression reduced as (x + y)⁵ + (x – y)⁵

Using binomial expansion, we have

(x + y)⁵ + (x – y)⁵ = x⁵ + ⁵C₁ x⁴y + ⁵C₂ x³y² + ⁵C₃ x²y³ + ⁵C₄ xy⁴ + ⁵C₅ y⁵ + x⁵ – ⁵C₁ x⁴y + ⁵C₂ x³y² – ⁵C₃ x²y³ + ⁵C₄ xy⁴ – ⁵C₅ y⁵

= 2(x⁵ + 10 x³y² + 5xy⁴)

= 2(x⁵ + 10 x³(x³ – 1 ) + 5x (x³ – 1)²)

= 2{10x⁶ + 5x⁷ + x⁵ – 10x³ – 10x⁴ + 5x}

Sum of coefficient of odd powers of x: 2{5 + 1 – 10 + 5} = 2

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